Finding Acceleration of a Ball in Constant Motion

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Homework Help Overview

The problem involves a ball that is initially at rest and accelerates uniformly, covering a distance of 10 meters during its third second of motion. The subject area pertains to kinematics and the equations of motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate equations to use for calculating distance covered in each second, with some suggesting the use of displacement formulas. There is an exploration of how to find the acceleration based on the distances covered during specific time intervals.

Discussion Status

Some participants have provided hints and guidance on how to approach the problem, including suggestions to calculate distances after specific time intervals and to find the difference in distances to solve for acceleration. Multiple interpretations of the problem are being explored, but there is no explicit consensus on the final approach.

Contextual Notes

Participants are working under the assumption that the ball starts from rest and are attempting to derive the acceleration based on the distance covered in the third second of motion.

ride4life
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Homework Statement


A ball initially at rest and constantly accelerating covers a distance of 10 metres during its third second of motion. What is its acceleration?


Homework Equations


No idea which equations I need. I'm guessing I need
a = (v-u)/t

The Attempt at a Solution


I've got no idea how to even apporach this. I'm thinking:
s = 3 s
u = 0 m/s
v =
a =
t =
 
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Welcome to PF!

Hi ride4life! Welcome to PF! :smile:

Hint: call the acceleration a, and find the distances after 2 seconds and after 3 seconds, and find the value of a that makes their difference equal 10 metres :wink:
 
What formula should I use to figure the distanes covered in each second?

0 - 1 = ?
1 - 2 = ?
2 - 3 = 10m = 10m/s/s

I was thinking in the 1st second it traveled 2.5m and in the 2nd second it traveled 5m. I think that turns out to be 2.5m/s/s. That's just a guess. That's all I can figure out. :(
 
You can use, displacement s = ut +1/2*a*t^2.
Find s for 2s and 3s. Then find the difference which is equal to 10m.
 
So for
2 - 3 = 10m

s = 10m
u = ?
a = ?
t = 1s
 
Ball starts from rest. So u = 0.
S2 = 1/2*a*(2)^2...(1)
S3 = 1/2*a*(3)^2...(2)
Find S3 - S2, equate it to 10m, and solve for a.
 
Thanks for the help, I got 4m/s/s which is the right answer. :D
 

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